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Math Help - Initial Value Problems

  1. #1
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    Initial Value Problems

    It's been awhile since I have had calculus and dealing with differential equations and initial value problems. Looking for some help with some easy problems.

    Find the solution of the given initial value problem.

    (1) y' + 2y = te^{-2t}, y(1)=0

    (2) t^3y' + 4t^2y = e^{-t}, y(-1)=0, t < 0

    A walkthrough of these would be greatly appreciated. Thanks in advance
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  2. #2
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    Quote Originally Posted by Proof_of_life View Post
    (1) y' + 2y = te^{-2t}, y(1)=0
    The integrating factor is f(x) = e^{\int 2 dx}= e^{2x}.
    Therefore the solution is,
    y=\frac{1}{f(x)} \int f(x) xe^{-2x} dx = e^{-2x} \int x dx = \frac{1}{2}x^2e^{-2x} + ke^{-2x}.
    Now y(1) = \frac{1}{2}e^{-2} + k e^{-2} = 0 \implies k = -\frac{1}{2}.
    Thus, y = \frac{1}{2}x^2e^{-2x} - \frac{1}{2} e^{-2x}.

    (2) t^3y' + 4t^2y = e^{-t}, y(-1)=0, t < 0

    A walkthrough of these would be greatly appreciated. Thanks in advance
    This one is similar. Just divide out the leading coefficient first.
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  3. #3
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    I was about to reply that I figured out both of these. Thanks for replying though!
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